Problem: Let $R$ be the region inside the polar curve $r=\cos(\theta)$ and inside the polar curve $r=1+\sin(\theta)$, as shown in the graph. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{2} \int_{0}^{\pi}\left(\cos^2(\theta)-\left(1+\sin(\theta)\right)^2\right)d\theta$ (Choice B) B $\dfrac{1}{2} \int_{0}^{\scriptsize\dfrac{\pi}{2}}\cos^2(\theta)\,d\theta+\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{2}}^{0}\left(1+\sin(\theta)\right)^2d\theta$ (Choice C) C $\dfrac{1}{2} \int_{0}^{\scriptsize\dfrac{\pi}{2}}\cos^2(\theta)\,d\theta+\dfrac{1}{2} \int_{\scriptsize\dfrac{\pi}{2}}^{\pi}\left(1+\sin(\theta)\right)^2d\theta$ (Choice D) D $\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{\pi}{2}}\left(\cos^2(\theta)-\left(1+\sin(\theta)\right)^2\right)d\theta$
Explanation: Since we are dealing with two separate polar curves, a good first step is to identify two areas, each enclosed by a single curve, that together define $R$. Such are $R_1$ and $R_2$ : $y$ $x$ $R_1$ $R_2$ $ 1$ $ 1$ $R_1$ is enclosed by $r=\cos(\theta)$ and $R_2$ is enclosed by $r=1+\sin(\theta)$. Once we express them as integrals, we can find $R$ using the following relationship: $\text{Area of }R=\text{Area of }R_1+\text{Area of }R_2$ $R_1$ is enclosed by $r=\cos(\theta)$ between $\alpha=0$ and $\beta=\dfrac{\pi}{2}$ : $\text{Area of }R_1=\dfrac{1}{2} \int_{0}^{\scriptsize\dfrac{\pi}{2}}\cos^2(\theta)\,d\theta$ $R_2$ is enclosed by $r=1+\sin(\theta)$ between $\alpha=-\dfrac{\pi}{2}$ and $\beta=0$ : $\text{Area of }R_2=\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{2}}^{0}\left(1+\sin(\theta)\right)^2d\theta$ Now we can express the area of $R$ : $\begin{aligned} &\phantom{=}\text{Area of }R \\\\ &=\text{Area of }R_1+\text{Area of }R_2 \\\\ &=\dfrac{1}{2} \int_{0}^{\scriptsize\dfrac{\pi}{2}}\cos^2(\theta)\,d\theta+\dfrac{1}{2} \int_{-\scriptsize\dfrac{\pi}{2}}^{0}\left(1+\sin(\theta)\right)^2d\theta \end{aligned}$